\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=\dfrac{1}{a+b+c}\)
\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)=abc\)
\(\Leftrightarrow a^2b+ab^2+abc+abc+b^2c+bc^2+a^2c+abc+ac^2-abc=0\)
\(\Leftrightarrow ab\left(a+b+c\right)+bc\left(a+b+c\right)+ac\left(a+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(ab+bc\right)+ac\left(a+c\right)=0\)
\(\Leftrightarrow\left(a+c\right)\left(ab+b^2+bc+ac\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
\(\circledast Với:a=-b\) , ta có :
\(P=\left(-b+b\right)\left(b^3+c^3\right)\left(c^5+a^5\right)=0\)
\(\circledast Với:b=-c\) , ta có :
\(P=\left(a+b\right)\left(b^3-b^3\right)\left(c^5+a^5\right)=0\)
\(\circledast Với:c=-a\) , ta có :
\(P=\left(a+b\right)\left(b^3+c^3\right)\left(-a^5+a^5\right)=0\)
KL..............
